proceedings of the american mathematical society Volume 32, Number 1, March 1972 CONJUGACY SEPARABILITY OF GROUPS OF INTEGER MATRICES PETER F. STEBE Abstract. An element £• of a group G is conjugacy distinguished if and only if given any element h of G either g is conjugate to h or there is a homomorphism £ of G onto a finite group such that f (g) is not conjugate to |(A). Following A. W. Mostowski, a group is conjugacy separable if every one of its elements is conjugacy dis- tinguished. Let GL(«, Z) be the group of n x n integer matrices with determinant ±1. Let SL(«, Z) be the subgroup of GL(«, Z) con- sisting of matrices with determinant +1. It is shown that GL(n, Z) and SL(n, Z) are conjugacy separable if and only if n=\ or 2. The groups SL(«, Z) are also called unimodular groups. Let GL(h, Zp) be the group of invertible />-adic integer matrices and SL(», Z„) be the group of /?-adic integer matrices with determinant 1. It is shown that GL(n, Z„) and SL(n, Z„) are conjugacy separable for all n and all/». 1. Introduction. A. W. Mostowski [4] defined conjugacy separable groups (see the abstract to this paper) and showed that the conjugacy problem is solvable in finitely presented conjugacy separable groups. It has been shown [6] that the free products of conjugacy separable groups are conjugacy separable and the elements of infinite order in a finite ex- tension of a free group are conjugacy distinguished: According to H. S. M. Coxeter and W. O. J. Moser [2, p. 85], the group GL(2, Z) has the presentation (x, y, z; x2=y2=z2=\, (xyf=(xz)2, (*z)4=l). Clearly GL(2, Z) is the free product of the groups (71=(x, y; x2=y2=l, (xy)<s=\) and G2=(v, z;v2=z2=l, (vz)*=l) with amalga- mating relations x=v and (vz)2= (xy)3. Thus an abelian subgroup of order 4 is amalgamated. The group SL(2, Z) is a subgroup of index 2 in GL(2, Z) and has the presentation (x,y\ x2=y3, jc*=1). These presentations will be used to show that GL(2, Z) and SL(2, Z) are conjugacy separable. 2. Conjugacy separability of GL(2, Z) and SL(2, Z). Theorem 1. The group GL(2, Z) is conjugacy separable. Received by the editors May 6, 1971. AMS 1970 subject classifications. Primary 20H05, 20E25. Key words and phrases. Group, unimodular group, conjugacy problem, conjugacy separable. © American Maihematical Society 1972 1 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2 P. F. STEBE [March Proof. By the remarks in the Introduction, there is a free group F such that [SL(2,Z); F]<<x>and [GL(2,Z):SL(2,Z)]<oo.Thus [GL(2,Z):F]< oo. According to [6, Theorem 2], every element of infinite order in GL(2, Z) is conjugacy distinguished in GL(2, Z). It follows from [3, Corollary 4.9.1] that the elements of finite order in GL(2, Z) are conjugate to elements of the factors Gi and G2 described in the Introduction. Thus, to show that GL(2, Z) is conjugacy separable we need only show that the conjugates of elements of Gj and G2 are conjugacy distinguished. Let g be an element of GL(2, Z) conjugate to an element of Gi or G2. Let h be any element of GL(2, Z) not conjugate to g. If h has infinite order in GL(2, Z), h is conjugacy distinguished in GL(2, Z) so there is a homomorphism f of GL(2,Z) onto a finite group such that £(g) is not conjugate to Ç(h) in £(GL(2,Z)). Thus we need only consider h of finite order in GL(2,Z) and hence h conjugate to an element of Gx or G2. Clearly, to show that there is a homomorphism f of GL(2,Z) onto a finite group such that |(g) is not conjugate to i(h) in GL(2, Z) we can replace g and h by their conjugates in G\ or G2, and by representatives of their conjugacy classes in these subgroups. The elements 1, x, y, xy, (xy)2 and (xy)3 are a com- plete set of conjugacy class representatives for the subgroup Gv Note that the defining relation (xy)3=(xz)2 implies that yxyxy=zxz. Since x,y and zare of order 2, x is conjugate to y in GL(2, Z). Also, the elements l,v,z, vz and (uz)2 are a complete set of conjugacy class representatives for the subgroup G2. Using the identifications x=v and (uz)2=(xy)3 we conclude that every element of finite order in GL(2, Z) is conjugate to one of the elements of the set {1, x, z, xz, (xz)2, xy, (xy)2}. The orders of those elements are, respectively {1, 2, 2, 4, 2, 6, 3}. If r] is a finite representation of GL(2, Z) faithful on the factors Gx and G2 of GL(2, Z), the images of two elements of different order will not be conjugate in r;(GL(2, Z)). According to B. H. Neumann [5, p. 532], such a representation exists. Thus we need only consider g and h conjugate to different elements of the set (x, z, (xz)2). Let f be the representation of GL(2, Z) induced by imposing the relation y=x. The image of GL(2, Z) is generated by u=r¡(x), w=r¡(z) with relations w2= h'2=(«m')2=1. Clearly r¡(x)rfjr¡(z), r¡(x)rijr¡((xz)2)=\ and í?(z)<-/^?((xz)2)=1. Theorem 2. The group SL(2, Z) is conjugacy separable. Proof. Since SL(2, Z) has the presentation (x, y; x2=y3, x*=l), it is the free product of a cyclic group of order 4 and a cyclic group of order 6 with amalgamation. Every element of finite order in SL(2, Z) is conjugate to an element of a factor of SL(2, Z), so that an element of finite order in SL(2, Z) is conjugate to a power of x or y. Let r¡ be the homomorphism of License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1972) CONJUGACY SEPARABILITY OF GROUPS 3 SL(2,Z) onto the cyclic group of order 12 (u; u12=\) given by r¡(x)=u3, r¡(y)=u2. The conjugacy class representatives of the elements of finite order in SL(2, Z) are the elements (1, x, x2, x3, y, y2, y*, yb). Their r\ images are, respectively, (1, u3, u6, u9, u2, w4, w8, m10). Thus if g and h are any two elements of finite order in SL(2, Z), either g is conjugate to h or r¡(g) is not conjugate to r¡(h). Let g and h be any two nonconjugate elements of SL(2, Z). Since SL(2, Z) has a free subgroup of finite index, every element of infinite order in SL(2,Z) is conjugacy distinguished. Hence to prove conjugacy separability, we may assume that g and h are of finite order. Then r¡(g) is not conjugate to r¡(h), so g is conjugacy distinguished. Hence SL(2, Z) is conjugacy separable. 3. The groups GL(n, Z) and SL(«, Z). Let A and B be the matrices 17(11)+1 25(11) " A = ll2 16(11)+ lj' 17(11) +1 11 B = 25(H)2 16(11)4-1.' Example 1. The matrices A and B have the following properties: (i) determinant A=determinant B=\; (ii) neither A nor B has eigenvalue 1 ; (iii) if n is an integer there is an integer matrix Tn such that TnA = BTn mod (n) and determinant Tn= 1 ; (iv) there is no 2 x 2 integer matrix Jsuchthat TA=BTand determinant T-±\. Argument. Properties (i) and (ii) follow from a simple computation. To obtain (iii) we need a lemma. Lemma 1. Let T be a 2x2 integer matrix. Let n be an integer. If determinant T=\ mod n there is an integer matrix Usuch that determinant U= 1 and U= T mod n. Proof. Let T=(ttj), i=\, 2,j=l, 2. Let d be the greatest common divisor of tn and ?12.Let tu—t*id, ti2=t*2d, so that t*t and t*2 are relatively prime integers. Thus there are integers a and b such that at*2— bt*1= 1. Let determinant T=\+rn. Let U be the matrix ~tu + n(a 4- ctu) tn + n(b + ci12)~ . t2i + n dt*i t22 -f n dt*2 . with c=bt21—at22—r, d=—cr. Clearly £7=7 mod n and it follows from evaluation that determinant U=l. The matrix U was suggested by Edward A. Bender. Lemma 1 implies that (iii) is shown if we can show that for each n there License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 4 P. F. STEBE [March is a matrix Tn such that TnA = BTn mod n and determinant Tn= 1 mod n . By the Chinese Remainder Theorem, we can restrict our attention to « a power of a prime p. Let V(x, y) be the polynomial matrix x y .11y 25* —y_ By a computation we obtain V(x,y)A—BV(x,y).